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First part asks for the probability of rolling an even number. Second part asks for the probability of not rolling either of two given numbers.

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For equally likely outcomes, you can calculate the probability of a particular event occurring by using the formula

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$\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$.

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Rolling a fair six-sided die has six possible outcomes, each of which is equally likely.

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Let's say we want to find the probability of rolling a $2$. There is only one outcome which involves a $2$ being rolled, so the number of favourable outcomes is $1$.

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Hence using the above formula,

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\\begin{align}
P(\\text{rolling a $2$}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{1}{6}
\\end{align}

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#### a)

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There are three possible outcomes where we roll an even number on the die:

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• we roll a $2$;
• \n
• we roll a $4$;
• \n
• we roll a $6$.
• \n
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Using the formula for probability for equally likely outcomes, this means that

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\$P(\\text{rolling an even number}) = \\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}= \\frac{3}{6} = \\frac{1}{2} \$

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#### b)

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To find the probability of not rolling a $\\var{die1}$ or a $\\var{die2}$, we use the same formula again.

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The total number of outcomes is still $6$.

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Here, we have four possible outcomes which don't involve rolling a $\\var{die1}$ or a $\\var{die2}$, i.e. when we roll any of the other numbers on the die.

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Using the formula,

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\$P(\\text{not rolling a \\var{die1} or a \\var{die2}}) = \\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}} = \\frac{4}{6} = \\frac{2}{3} \$

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You're going to roll a fair six-sided die.

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number of red balls in part c

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Not included number for a) ii)

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Not included number for a) ii)

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What is the probability of rolling an even number?

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What is the probability of not rolling a $\\var{die1}$ or $\\var{die2}$?

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